Chem 127 Monday1 Nov
Exams – graded Wed
Web site, updated this morning
Paper and topic selection (add topics this afternoon)
Topic – Kinetics or the Rate
of Reactions chapter
Problems
from end of chapter (post today) Friday’s lab Work
Summary – focus on the rate at which
reactions occur
• For
now, only in homogeneous (uniform) mixtures
•
Express rate as {change in concentration) / {time}
• A
convenient reference is half life
*time for half of the reactant to vanish
• (if several
species, probably the limiting reagent)
•
Can vary from nanosec to millions of years
• rate
usually depends on one or more of these
* temperature (all reactions faster hot than cold)
*old friend, plot of ln (rate) vs 1/T(K)
* concentration of one or more reactants
*may surprise you – rate could be independent
*role of catalyst (faster, but not a
reactant)
* rarely, concentration of product
* rarely, inhibited by another species
___________
Let’s Ignore Temperature for now
until stated otherwise, temperature is kept the same
many reactions
follow a simple form ...
Rate = (k) * [reactant n]' an n-th order reaction
k is a constant, the rate constant for this reaction
or Rate = k *
[reactant A]a * [species B]b ...
seldom more than two terms, rarely more than 3
reaction is a-th order in A and b-th order
in B
[Species] could be reactant or catalyst
The rate and the rate constant are always
treated as positive values.
Simplest case is a
true zero-th order reaction
Rate of the reaction doesn’t depend on concentration
At least over a very wide concentration range –
obviously no reactant – - no reaction
Pretty simple – rate stays constant
whatever change occurs in first second,
minute
same in any following sec, minute
plot of [A] vs. time is straight line
down if A= reactant
up if A = product
of course, stopping when we run out of
reactant
Why zeroth order?
___________
Simplest explanation – some other factor fixes the rate
In Friday’s lab – cyanotype or blue print
amount of blue proportional to
a. time of exposure
b. intensity (and location) of light source but virtually independent on how much is left
The electroforming
we tried – Cu" – - > Cu metal
rate is independent of [Cu"]
Proportional to electrical current
Many biological
reactions (metabolism)
rate is independent of the material that reacts
enzyme reactions – they do depend on amount of
enzyme, so not true zeroth order...
but
we might measure reaction without identifying enzyme’s role.
Simple test – Plot [Reactant] vs time
If straight line, it’s zeroth order
But plot needs to extend beyond 1 half life
(gentle curve looks like straight line in any short segment)
Slope
is the rate constant
(In
this case, works just as well to plot [Product] )
Another test –
Prepare several reaction mixtures
measure how much reacts over a short period of time
usually <10% of a half life
if rate is unchanged, it’s zeroth order in (that) reactant
___________
More Common is First Order Reaction
Rate = k [ Reactant]
here
there’s either one reactant
or
reaction is zeroth order in other reactants
Clearly
as the reaction proceeds
[A] decreases and Rate decreases
Simple test –
plot of [A] vs time shows typical form (exponential)
but that’s hard to confirm visually
Plot of 1n [A] vs time shows a straight line
slope is the rate constant (negative
slope, positive k)
again, usually need to plot beyond 1 half life
notice – doesn’t work with [product]
In lab work, it’s often easier to measure product need to use chemical
equation
compute the remaining reactant
Another very useful rule and test
T 1/2 is time for conc. of A to
drop to 1/2 [Ao]
in same time, it will go from 1/2 [Ao] to 1/4
[Ao]
no other reaction follows that pattern
after n (T 1/2) periods we have
½, 1/4, 1/8, 1/16, 1/32, 1/64, 1/128
(1/2)10 is useful (1/2)'º = 1/1024 or about 1/1000
Can Also use Initial
Rate as,a test
___________
prepare mixtures of different concentrations of A
let about 5% react
measure [A] before and after
compute rate delta[A] / delta-time
for first order reaction, this rate is proportional to conc
double [A] and rate doubles
cut [A] in half, reaction is half as fast
any other conditions (temp, conc of others) held constant
why the 5% rule?
*since the rate changes as [A] is used, the reaction
does not have a single rate
*however, over first 5%, that effect is small (<5%)
*why not first 0.1%? (works, but might not be as
easy to measure the change in concentration)
Let’s look at a common biological problem
Glucose+ 02 – – > (products)
very slow unless (a) concentrated NaOH <blue bottle demo>
(b) enzyme present (glueose oxidase)
Rate = k [Glucose]' [02]? [enzyme]'
we can use the rate of reaction
to measure blood (or urine) glucose level
(diabetics need to monitor this)
first... [O2] is very constant if equilibrium with air
and [O2] decreases very little due to this reaction
We can provide a
constant enzyme concentration in test supplies
usually bound to a solid (test strip or testing electrode)
notice that [enzyme] does not vary during reaction (catalyst)
So reaction is First order in glucose (everything else is controlled)
___________ __ But how do you measure [gluocse] to determine the rate and if you can determine the glucose, why bother with rate?
In this case [glucose] + 02 + (enzyme) – > prodcut + H2O2
there’s a very good electrode for measuring
the H2O2
teststrips, use H2O2 as
bleach
mix with a dye, and match the rate of color
loss
One important
general principle
laws based on plots (amounts remaining over time)
need to treat enzyme differently
initial rate method
finds the order with respect to enzymes
easily
rate = k [A]‘
plot of 1/[A] vs. time is straight line
¿slope¿
is rate constant again, over 1-3 half lives
Half life rule is different (useful)
half vanishes in t>¿¿
it takes twice as long to go from % [Ao) to
1/4 [Ao]
it takes four time that interval to go from
'/4 to 1/8 [Ao]
The initial rate method works well
make test runs with varying [A],
if second order, rate 4x faster if you double
[A]
again, 5%
rule
This also works in a different but common case (but not universal)
reaction is first order in [A] and [B)
reaction is 1:1
start with [Ao] = [Bo]
___________